Focal length of concave lens using convex lens

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SUMMARY

The discussion focuses on calculating the focal length of a concave lens when used in conjunction with a convex lens. A convex lens with a focal length of 10 cm forms a real image at 20 cm from the object. When a concave lens is introduced midway, the image shifts to 5 cm from its original position. The calculations indicate that the focal length of the concave lens is -30 cm, but the expected answer is -5 cm, highlighting a potential error in the image distance assumption for the concave lens.

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  • Knowledge of real and virtual image formation in optics.
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  • Review the lens formula for concave lenses and its application in image formation.
  • Study the concept of virtual objects in optics and their implications on calculations.
  • Explore the differences between real and virtual images produced by convex and concave lenses.
  • Practice problems involving multiple lenses to solidify understanding of focal lengths and image distances.
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Amith2006
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# A convex lens of focal length 10 cm forms a real image of an object placed at a distance of 20 cm from it. Midway between the convex lens and the position of the image a thin concave lens is introduced. The image formed now is at a distance of 5 cm away from the earlier position. What is the focal length of the concave lens?
I solved it in the following way:
Case-1
Let u, v and f be the object distance, image distance and the focal length of the convex lens.
Here u = 20 cm, f = + 10 cm
1/v = 1/f – 1/u
1/v= 1/10 – 1/20
v = 20 cm
Case- 2
Let u, v and f be the object distance, image distance and the focal length of the concave lens.
Here u = - 10(virtual object), v = + 15 cm
1/f = -1/10 + 1/15
f = - 30 cm
But the answer given in my book is – 5 cm.
 
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I don't get 5 cm either.

But I think you made a mistake with setting v=15cm in the second part. Would you expect the image for a concave lens to be 5cm closer to the lens or 5cm farther from it (compared to the object)?
 
I assumed it to be away because i felt that concave lens is diverging in nature and so the image would be produced 5 cm away from the lens. What do you say?
 
That sounds right (that the image would be farther away). There must be a mistake with the way I was doing it. I'm too sleepy to think straight now. I'll look at this again when I find some free time. Meanwhile someone else might have something useful to say.
 

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